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Natalija [7]
3 years ago
6

What’s powerd by energy from earths intorior

Chemistry
2 answers:
Travka [436]3 years ago
6 0

Erosion is the final answer :)

Grace [21]3 years ago
5 0

Whats the choices.?.?.?



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It is proposed to use Liquid Petroleum Gas (LPG) to fuel spark-ignition engines. A typical sample of the fuel on a volume basis
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Answer:

a)

The overall  balanced combustion  reaction is written as :

0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> 3.05CO_2  \ + \ 3.8H_2O \ + \ 18.612N_2

(F/A)_{stoichiometric} = 0.0424

(A/F)_{stoichiometric} = 23.562

b)

the higher heating values (HHV)_f per unit mass of LPG = 49.9876 MJ/kg

the lower heating values (LHV)_f per unit mass of LPG = 46.4933 MJ/kg

Explanation:

a)

The stoichiometric equation can be expressed as :

0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> aCO_2  \ + \ bH_2O \ + \ cN_2

Now, equating the coefficient of carbon; we have:

(0.7×3)+(0.05×4)+(0.25×3) = a

a = 3.05

Also, Equating the coefficient of hydrogen : we have:

(0.7 × 8) +(0.05 × 10) + ( 0.25 × 6) = 2 b

2b = 7.6

b = 3.8

Equating the coefficient of oxygen

2x = 2a + b

x = \frac{2a+b}{2} \\ \\ x =  \frac{2(3.05)+3.8}{2} \\ \\ x = 4.95

Equating the coefficient of Nitrogen

c = 3.76x \\ \\ c = 3.76 *4.95 \\ \\ c = 18.612

Therefore, The overall  balanced combustion  reaction can now be written as :

0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> 3.05CO_2  \ + \ 3.8H_2O \ + \ 18.612N_2

Now;  To determine the stoichiometric F/A and A/F ratios; we have:

(F/A)_{stoichiometric} = \frac{n_f}{n_a } \\ \\  (F/A)_{stoichiometric} = \frac{1}{4.95*(1+3.76)} \\ \\ (F/A)_{stoichiometric} = 0.0424

(A/F)_{stoichiometric} = \frac{n_a}{n_f } \\ \\  (A/F)_{stoichiometric} = \frac{4.95*(1+3.76)}{1} \\ \\ (A/F)_{stoichiometric} = 23.562

b)

What are the higher and lower heating values per unit mass of LPG?

Let calculate the molecular mass of the fuel in order to determine their mass fraction of the fuel components.

Molecular mass of the fuel M_f = (0.7*M_{C_3H_5} ) + (0.05 *M_{C_4H_{10}}) + (0.25*M _{C_3H_6})

= 30.8 + 2.9 + 10.5

= 44.2 kg/mol

Mass fraction of the fuel components can now be calculated as :

m_{C_3H_8} = \frac{30.8}{44.2} \\ \\ m_{C_3H_8}  = 0.7 \\ \\ \\  m_{C_4H_{10}} = \frac[2.9}{44.2} \\ \\ m_{C_4H_{10}} = 0.06  \\  \\ \\ m_{C_3H_6} = \frac{10.5}{44.2} \\ \\ m_{C_3H_6}  = 0.24

Finally; calculating the higher heating values (HHV)_f per unit mass of LPG; we have:

(HHV)_f=(0.7 * HHV_{C_3H_8}) + (0.06 *HHV_{C_4H_{10}})+(0.24*HHV_{C_3H_6} \\ \\ (HHV)_f=(0.7*50.38)+(0.06*49.56)+(0.24*48.95) \\ \\ (HHV)_f=49.9876 \ MJ/kg

calculating the lower heating values (LHV)_f per unit mass of LPG; we have:

(LHV)_f = (HHV)_f - \delta H_w \\ \\  (LHV)_f = (HHV)_f  - [\frac{m_w}{m_f}h_{vap}] \\ \\ (LHV)_f   = 49.9876 \ MJ/kg -  [\frac{3.8*18}{44.2}*2.258 \ MJ/kg]  \\ \\ (LHV)_f = 46.4933 \ M/kg

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